3.5 Harmonic Functions

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We determine and create harmonic functions.

A function $u(x,y)$ is called harmonic on a disk $D \subset \R ^2$ if the second order partial derivatives $u_{xx}$ and $u_{yy}$ exist on $D$ and satisfy the equation $\Delta u = u_{xx} + u_{yy} = 0$ on $D$ .

example 1 The function $u(x,y) = 3x + 5y +1$ is harmonic in the entire plane, $\R ^2$ since $u_{xx} \quad \text {and} \quad u_{yy} = 0$ everywhere. Hence $\Delta u = 0$ on $\R ^2$ and $u$ is harmonic everywhere.

(problem 1) Select all of the functions that are harmonic on $\R ^2$ ?

$u(x,y) = 2x -3y$ $u(x,y) = xy$ $u(x,y) = x^2 + y^2$ $u(x,y) = x^2 - y^2$

example 2 Show that the function $u(x,y) = \ln \left (x^2 + y^2\right )$ is harmonic in the punctured plane, $\R ^2 \backslash (0,0)$ .
Using the chain rule gives $u_x = \frac {2x}{x^2 + y^2}$ Next, the quotient rule gives $u_{xx} = \frac {(x^2+y^2)(2)- (2x)(2x)}{\left (x^2 +y^2\right )^2} = \frac {2y^2-2x^2}{\left (x^2 +y^2\right )^2}$ Due to the symmetry in variables in the formula for $u(x,y)$ , we can see that $u_{yy} = \frac {2x^2-2y^2}{\left (x^2 +y^2\right )^2}$ Adding these gives $\Delta u = u_{xx} + u_{yy} = 0$ since the numerators are opposites (and the denominators are equal). The above calculations are valid for all $(x,y)\in \R ^2$ except $(0,0)$ (where $u$ is not defined). Hence $u$ is harmonic in the punctured plane, $R^2\backslash (0,0)$ .

(problem 2) Show that the following functions are harmonic on the indicated set.
\begin{align*} a) \;\; u(x,y) &= e^x\cos y \;\;\text{on} \;\; \C \\ u_y &= \answer{-e^x \sin y}\\ u_{yy} &= \answer{-e^x \cos y}\\ b) \;\; u(x,y) &= \tan ^{-1}\left (\frac{y}{x}\right ) \;\; \text{on} \;\; \C \backslash \{y-\text{axis}\}\\ u_x &= \answer{-\frac{y}{x^2+y^2}}\\ u_{xx} &= \answer{\frac{2xy}{(x^2+y^2)^2}} \end{align*}

Recall $\frac {d}{dx} \tan ^{-1} x = \frac {1}{1+x^2}$

Simplify $u_x$ before calculating $u_{xx}$

Here is a video solution to problem 2b:

If $f(z) = u(x,y) + iv(x,y)$ is analytic in a disk $D$ , and if the second order partial derivatives of $u$ and $v$ exist and are continuous on $D$ then $u$ and $v$ are both harmonic on $D$ .

Proof: We prove that $u$ is harmonic on $D$ and leave the proof of $v$ as an exercise.
By the Cauchy-Riemann equations, $u_y = -v_x$ on $D$ . Therefore, $u_{yy} = -v_{xy} = -v_{yx}$ where the second equality follows from the symmetry of mixed partial derivatives. Applying the Cauchy-Riemann equations (specifically $v_y = u_x$ ) gives $u_{yy} = -v_{yx} = -u_{xx}$ on $D$ . Hence $\Delta u = u_{xx} + u_{yy} = u_{xx} - u_{xx} = 0$ and $u(x,y)$ is harmonic on $D$ . $\blacksquare$

(harmonic) Suppose $f(z) = u(x,y) + iv(x,y)$ satisfies the hypotheses of the above theorem on a disk $D$ . Prove $v$ is harmonic on $D$ .
To begin, note that $v$ is the real part of which function?

$-f(z)$ $i f(z)$ $-i f(z)$

Is this function analytic on $D$ ? yesno
How do we know $v$ is harmonic on $D$ ? direct computation of $\Delta v$ the theorem above

Suppose $u(x,y)$ is harmonic on a disk $D$ . Then there exists a complex function $f(z)$ , analytic on $D$ , such that $u(x,y)$ is the real part of $f$ . The imaginary part of $f$ is called the harmonic conjugate of $u$ .

example 3 Find the harmonic conjugate of $u(x,y) = x^3 - 3xy^2$ .
First, we should verify that $u$ is harmonic: $\Delta u = u_{xx} + u_{yy} = (6x) + (-6x) = 0$ so $u$ is harmonic on $\C$ . According to the theorem we can find $v(x,y)$ such that $f(z) = u(x,y) + iv(x,y)$ is analytic (entire in this case). To construct $v$ , we begin by noting that $v_y = u_x = 3x^2 - 3y^2$ since $u$ and $v$ must satisfy the Cauchy-Riemann equations. Now, we integrate with respect to $y$ treating $x$ as a constant: $v(x,y) = \int \left (3x^2 -3y^2\right ) \; dy$ which yields $v(x,y) = 3x^2y - y^3 + C(x)$ Note that the constant of integration is written as $C(x)$ since we were assuming in the integral that $x$ was constant. To determine $C(x)$ , we use the other Cauchy-Riemann equation. On one hand, $v_x = -u_y = 6xy$ and on the other hand, since we have established a form for $v$ , we can use it to obtain $v_x = 6xy + C'(x)$ . Equating these two expressions for $v_x$ we see that $C'(x) =0$ and hence $C(x)$ is a constant. Thus the harmonic conjugate of $u(x,y) = x^3 - 3xy^2$ is $v(x,y) = 3x^2y - y^3 + C$ where $C$ is any real constant.

In general, a harmonic function $u$ has infinitely many harmonic conjugates, each pair differing by a constant.

(problem 3) Verify that each function is harmonic on $\C$ and then find its harmonic conjugate. \begin{align*} a) \;\;u(x,y) &= x^4 -6x^2y^2 + y^4\\ v_y &= \answer{4x^3 - 12xy^2}\\ v(x,y) &= \answer{4x^3y - 4xy^3}+C\\[8pt] b) \;\;u(x,y) &= e^x\sin y\\ v_y &= \answer{e^x \sin y}\\ v(x,y) &= \answer{-e^x \cos y}+C\\[8pt] c) \;\;u(x,y) &= \cos x \cosh y\\ v_y &= \answer{-\sin x \cosh y}\\ v(x,y) &= \answer{-\sin x \sinh y}+C\\[8pt] d) \;\; u(x,y) &= \sinh x \sin y\\ v_y &= \answer{\cosh x \sin y}\\ v(x,y) &= \answer{-\cosh x \cos y}+C \end{align*}

Here is a video solution to problem 3d:

2024-09-27 14:07:30